8 Answers
8

You want $x^{m+n}=x^mx^n$ to hold not only for positive integers $m,n$, but for all integers. The case $m=0$ leads us to $x^n=x^0x^n$, i.e. $x^0=1$, and the case $m+n=0$ then to $1=x^{-n}x^n$, i.e. $x^{-n}=1/x^n$.

Remark $\ $ Said slightly more technically, both $\,x^{-n}\,$ and $\,1/x^n\,$ are inverses of $\,x^n,\,$ so they are equal by uniqueness of inverses (the above proof is just a special case of the general proof of uniqueness of inverses). But this leads to the question as two why $\,x^n x^{-n} = 1.\,$ That's a special case of the law of exponents $\, x^{n+k} = x^n x^k,$ and the definition of negative powers is defined precisely to preserve this law when we extend to negative exponents (analogous to how operations on negative integers are uniquely inferred by the requirement that they should preserving useful properties or laws of positive integers, e.g. the distributive law). These preservation constraints are better understood as structure preservation when one studies algebraic sttructures in abstract algebra (e.g. above there is innate group theoretic structure).

by definition we write the product of $n$ elements $x$, say, as $x^n$. here $n$ is initially a positive integer. with this definition it follows immediately that for positive integers $m$ and $n$ we have the initial form of the first law of indices:

$$ x^mx^n = x^{m+n}
$$
it then follows that it is natural to interpret $x^0$ as $1$, since the following makes obvious sense
$$ x^mx^0 = x^{m+0}=x^m
$$
at this stage we have a natural isomorphism of the non-negative powers of an indeterminate, $x$, with the additive monoid $\mathbb{N}_{\{+\}}$

we know that this monoid can be located as a submonoid of the additive group $\mathbb{Z}_{\{+\}}$ of all the integers. in fact you can see that the structure of $\mathbb{Z}_{\{+\}}$ is naturally described as the factor monoid $\mathbb{N} \times \mathbb{N} /E$ where $E$ is the equivalence relation defined by:
$$
(m,n) \equiv (m',n') \; \text{iff} \; m+n'=m'+n
$$
as yoknapatawpha points out, we can pull back this group structure into the index notation, which, for $n \in \mathbb{Z}, n \lt 0$, gives us an extended set of "negative powers" of the indeterminate $x$.

when we interpret $x$ as a non-zero element of a multiplicative abelian group, the term subtraction is replaced by the term division. but note that both subtraction and division are non-associative operations. however the manipulation of operations including subtraction is made easier because of the rule:
$$ a - (b - c) = a + c -b
$$
and there is an exactly corresponding law for the operation of division:
$$ a(bc^{-1})^{-1} = acb^{-1}
$$

By definition, we have $x^{-1} = \frac{1}{x}$. Taking this to the $y$-th power on each side, we have $(x^{-1})^y = \frac{1^y}{x^y}$. Note that when we raise $x$ to an exponent twice, we multiply the two exponents -- to make this clear, think of $(x^2)^2 = x^2\cdot x^2 = x\cdot x \cdot x \cdot x = x^4 = x^{2\cdot 2}$ . This gives $x^{-y} = \frac{1}{x^y}$.

Because we have $x^y*x^z=x^{y+z}$ now we have $x^0=1$ and we have $\dfrac{x^y}{x^y}=1$ so $x^y*\dfrac{1}{x^y}=1$ So if we where to express $\dfrac{1}{x^y}$ as a power of x the sum of the exponents would have to be 0. so $\dfrac{1}{x^y}=x^{-y}$

We know that $\frac{1}{x} \times x = 1$ by using the factorization rule. Then by multiplying both side of the equation with $x^{-1}$ from the right hand side we will have $\frac{1}{x} = x^{-1}$. Then by applying this to your question, we have:

Where the last results came from multiplication rule for exponential; i.e. $x^{y_1} \times x^{y_2} = x^{y_1 + y_2}$ and to extend it to the case of having many $n$ exponential powers $x^{y_1} \times \cdots \times x^{y_n} = x^{y_1 + \cdots + y_n}$ by using mathematical Induction.

$\frac{2^1}{2^2}$ = $2^{1-2}$. Solving the exponents, we have $2^{-1}$. Reversing this, we get back to $\frac{2^1}{2^2}$, which, when expanded, gives us $\frac{2}{2 * 2}$. By distributive law, this becomes $\frac{2}{2} * \frac{1}{2}$, where the first operation gives us 1. This leaves us with $\frac{1}{2}$. Following that, variables or numbers with negative exponents should be reciprocals (i.e. $\frac{1}{n}$) since they can be solved the same way. Try it at home with

$\begingroup$Your statement $\dfrac{2^{1}}{2^{2}}=2^1-2^2$ should be corrected... It is not true. The statement you want is simply $\dfrac{2^{1}}{2^{2}}=2^{1-2}$.$\endgroup$
– Nick HDec 15 '14 at 23:01